blocks|key|4405368|text|您可以在xrange中使用step参数|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4405369|for+i+in+xrange(0,+len(html_doc),+2):
++++dict[html_doc[i]]+=+html_doc[i%2B1]|code-block|syntax|javascript|4405370|如果使用step，则xrange需要start参数。如果使用step，xrange将生成[start,+start%2Bstep,+start%2B2*step+...+]|4405371|entityMap^0|4|6|D|4|0|0|4|4|A|6|I|5|Z|6|18|12|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|N|$]]

you can use the <code>step</code> parameter in <code>xrange</code>:

<pre><code>for i in xrange(0, len(html_doc), 2):
 dict[html_doc[i]] = html_doc[i+1]
</code></pre>

<code>xrange</code> needs the <code>start</code> parameter if you use <code>step</code>. If you use step, <code>xrange</code> will generate <code>[start, start+step, start+2*step ... ]</code>

blocks|key|4405373|text|html_dict+=+dict(zip(html_doc[::2],+html_doc[1::2]))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4405374|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

blocks|key|2089123|text|将iter与dict和zip配对并从每两个元素创建键/值：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2089124|it+=+iter(html_doc)

print(dict(zip(it,it)))
{u'Buildings+Structures+And+Works+Exceeding+90+Metres':+u'Aerodrome+Safeguarding+London+Heathrow+1',+u'Flood+Zone+3+Low+Residual+Risk':+u'Flood+Zone+3+Low+Residual+Risk',+u"Embankment+Residents'+Association":+u"Embankment+Residents'+Assoc.",+u'Crabtree+Conservation+Area+-+Number+28':+u'Conservation+Area',+u'Controlled+Parking+Zone+T':+u'Controlled+Parking+Zone+T',+u'Environment+Agency+Flood+Risk+Zone+3':+u'Flood+Risk+Zone+3',+u'Environment+Agency+Flood+Risk+Zone+2':+u'Flood+Risk+Zone+2'}|code-block|syntax|javascript|2089125|如果对列表进行切片，则会创建两个新列表。|2089126|entityMap^0|1|4|6|4|B|3|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

Use <code>iter</code> with <code>dict</code> and <code>zip</code> to pair and create key/value from every two elements:

<pre><code>it = iter(html_doc)

print(dict(zip(it,it)))
{u'Buildings Structures And Works Exceeding 90 Metres': u'Aerodrome Safeguarding London Heathrow 1', u'Flood Zone 3 Low Residual Risk': u'Flood Zone 3 Low Residual Risk', u"Embankment Residents' Association": u"Embankment Residents' Assoc.", u'Crabtree Conservation Area - Number 28': u'Conservation Area', u'Controlled Parking Zone T': u'Controlled Parking Zone T', u'Environment Agency Flood Risk Zone 3': u'Flood Risk Zone 3', u'Environment Agency Flood Risk Zone 2': u'Flood Risk Zone 2'}
</code></pre>

If you slice the list you create two new lists.

blocks|key|1641188|text|我认为这是一种非常可读的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1641189|d+=+{}
while+html_doc:
++++d[html_doc.pop()]+=+html_doc.pop()|code-block|syntax|javascript|1641190|给予：|1641191|>>>+d
{u'Buildings+Structures+And+Works+Exceeding+90+Metres':+u'Aerodrome+Safeguarding+London+Heathrow+1',
+u'Controlled+Parking+Zone+T':+u'Controlled+Parking+Zone+T',
+u'Crabtree+Conservation+Area+-+Number+28':+u'Conservation+Area',
+u"Embankment+Residents'+Association":+u"Embankment+Residents'+Assoc.",
+u'Environment+Agency+Flood+Risk+Zone+2':+u'Flood+Risk+Zone+2',
+u'Environment+Agency+Flood+Risk+Zone+3':+u'Flood+Risk+Zone+3',
+u'Flood+Zone+3+Low+Residual+Risk':+u'Flood+Zone+3+Low+Residual+Risk'}|1641192|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I think this is a quite readable approach:

<pre><code>d = {}
while html_doc:
 d[html_doc.pop()] = html_doc.pop()
</code></pre>

Giving:

<pre><code>&gt;&gt;&gt; d
{u'Buildings Structures And Works Exceeding 90 Metres': u'Aerodrome Safeguarding London Heathrow 1',
 u'Controlled Parking Zone T': u'Controlled Parking Zone T',
 u'Crabtree Conservation Area - Number 28': u'Conservation Area',
 u"Embankment Residents' Association": u"Embankment Residents' Assoc.",
 u'Environment Agency Flood Risk Zone 2': u'Flood Risk Zone 2',
 u'Environment Agency Flood Risk Zone 3': u'Flood Risk Zone 3',
 u'Flood Zone 3 Low Residual Risk': u'Flood Zone 3 Low Residual Risk'}
</code></pre>

I have a list of items which has a length of 14:

<pre><code>html_doc = [u'Crabtree Conservation Area - Number 28',
u'Conservation Area',
u'Environment Agency Flood Risk Zone 3',
u'Flood Risk Zone 3',
u'Environment Agency Flood Risk Zone 2',
u'Flood Risk Zone 2',
u'Buildings Structures And Works Exceeding 90 Metres',
u'Aerodrome Safeguarding London Heathrow 1',
u'Controlled Parking Zone T',
u'Controlled Parking Zone T',
u'Flood Zone 3 Low Residual Risk',
u'Flood Zone 3 Low Residual Risk',
u"Embankment Residents' Association",
u"Embankment Residents' Assoc."]
</code></pre>

each of the elements in the array should be paired together as "Name":"Constraint_type" meaning that the first and second element of the list are related. 

I have this code which I don' think would work:

<pre><code>for i in xrange(len(html_doc)):
 dict.append("Name:" html_doc[i], "Constraint_Type": html_doc[i+1])
</code></pre>

Any Idea on this?

Building python dict out of a single list

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个长度为14的项目列表：html_doc = [u'Crabtree Conservation Area - Number 28',u'Conservation Area',u'Environment Agency Flood Risk Zone 3',u'Flood Risk Zone 3',u'Environ...

问从单个列表构建python字典
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问从单个列表构建python字典EN